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RUSSIAN OPEN SCHOOL ASTRONOMICAL OLYMPIAD BY CORRESPONDENCE
2008
PROBLEMS WITH SOLUTIONS
1. Problem. Observer is situated in the definite point on the
Earths surface. One definite moment he noticed that each point of
ecliptic had met the mysterious property: the angular distance
between this point and North Celestial Pole had been equal to the
zenith distance of the same ecliptic point. Disregarding the
refraction, please find the latitude of the observation point.
(O.S. Ugolnikov) 1. Solution. The conditions described above are
always met on the North Pole of the Earth (latitude +90) where the
North Celestial Pole coincides with the zenith and the angular
distance from the North Celestial Pole is equal to the zenith
distance not only for the points of the ecliptic, but for all
points of the celestial sphere. But this is not the only
solution.
If we observe somewhere far from the North Pole of the Earth,
than the North Celestial Pole and the zenith are two different
points of the celestial sphere. The problems conditions will be met
if these two points are symmetric relatively the ecliptic line. It
is seen in the figure for the example ecliptic point E.
Ecliptic
North Celestial Pole
SouthCelestial Pole
Horizon
Equator
Zenith
Nadir
E
Ecliptic is inclined to the equator by the angle equal to 23.4.
The angle between the northern polar direction (which is
perpendicular to the equator) and the ecliptic plane is equal
to
= 90 = 66.6.
Symmetry of North Celestial Pole and the zenith relatively the
ecliptic means that the angle between the zenith direction and the
ecliptic plane is the same. The ecliptic plane must be
perpendicular to the plane containing the zenith and northern polar
directions. Thus, the zenith distance of North Celestial Pole is
equal to
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Russian Open School Astronomical Olympiad by Correspondence
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zP = 2 = 133.2.
Northern Celestial Pole is below the horizon, and the
observation point is in the Southern hemisphere of the Earth. The
latitude is negative, its module is equal to the depth of Northern
Celestial Pole:
= (zP 90) = 90 2 = 90 + 2 = 43.2.
Finally, the problem condition can take place at the latitudes
+90 (it always takes place there) and 43.2. 2. Problem. The
artificial satellite of the Earth has the mass equal to 100 kg and
moves along the elliptical orbit with perigee altitude equal to 200
km and apogee altitude equal to 10000 km. Being close to perigee,
the satellite is decelerated by the Earths atmosphere. Please
estimate the time, during which the satellites orbit will become
circular. The decelerating force of the atmosphere can be
considered to be constant with the value 0.01 Newton, the path
length of the satellite through the atmosphere each revolution is
equal to the radius of the Earth. (O.S. Ugolnikov) 2. Solution.
Lets explain why the orbit of the satellite will turn to circle.
Being close to the apogee, the satellite is far from the Earth, it
is not decelerated by the atmosphere and moves by the elliptical
trajectory according to the Second Kepler law. Approaching the
perigee, the velocity of the satellite increases and exceeds the
circular velocity for this distance to the Earth. But here the
satellite is being decelerated by the atmosphere. As we will see
below, this deceleration is not too strong to lead the satellite to
fall down or burn up in the atmosphere during the first revolution.
But each revolution the satellite will loose the velocity and
energy before the escape from the atmosphere.
Earth
AtmosphereA
L
Let L will be the distance from the center of the Earth to the
upper border of the dense atmosphere layers, the atmosphere
deceleration above this border can be neglected. The satellite
velocity on this border is equal to
,122
=aL
GMv
where M is the Earth mass and a is orbit large semi-axis. Each
revolution the values of v and a are decreasing. While the orbit is
elliptical and satellite escapes from the atmosphere, its perigee
altitude is changing slowly. Thus, the orbit of the satellite turns
to the circle with the radius close to the perigee distance of the
initial orbit. When the satellite does not escape the atmosphere,
its trajectory will turn
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Russian Open School Astronomical Olympiad by Correspondence
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to spiral. The satellite will go down to the Earth increasing
its velocity until it burn or fall down. But this stage of orbit
evolution is not considered in this problem.
The following solution is the simplified method of the
calculation of the time of orbit evolution, that is quite
complicated problem in general. Lets consider two consecutive
moments of the satellites escape from the dense layers of
atmosphere (the point A in the figure). Of course, the escape
points are different, but their distances from the center of the
Earth, L, are the same. Lets draw the relations between the
velocities of the satellite at these moments and the values of
major semi-axes of the orbit at (i) and (i+1) revolutions:
,122
=
ii aL
GMv
,12
1
21
=
++
ii aL
GMv
According to the constant energy law,
,)( 22 1 DFvvm2 ii
=+ Here m is the satellite mass, F is the deceleration force, D
is the length of the satellite path through the atmosphere.
According to the problem condition, the values of F and D are
constant. Their multiplication is equal to 64 kJ, that is many
times less than the kinetic energy of the satellite in perigee.
Thus, the orbit will turn to circle slowly, during the large number
of revolutions. The change values of major semi-axis ai and orbital
period Ti are many times less than the values ai and Ti
themselves.
From the formulae above we obtain:
.2
;1122
21 GMm
aFDaa
aaaGMm
FD ii
i
i
ii==
+ Major semi-axis and orbital period are related with each other
by the III Kepler law:
.4 2
23
GMTa ii =
Mean change value of major semi-axis during (i)-revolution is
the value of ai divided by orbital period:
.1 GMmaFD
Ta ii
ii =
=
We can assume, that the change of major semi-axis in time will
have a power law, we can find the number n, for which:
.)(
ii
ni
in constTa ==
Our aim is to find the number n. The parameter in the numerator
of the last formula is equal to
.)()( 11+ +== niininiininini aanaaaaaa
Here we had used the mathematical property of small value :
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Russian Open School Astronomical Olympiad by Correspondence
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.1)1( nn ++
Finally, the change of value ain per unit of time is equal
to
.)( )21(1
GMmnaFDna
Ta
Ta nn
ii
i
i
ni
in ===
We see that if n=1/2 than this value will not depend on time.
Since the orbit evolution is much longer than the orbital period,
we can assume the major semi-axis decrease as continuous process.
The value of square root of major semi-axis will decrease by the
linear law:
.20
tGMm
FDaa =
Here a0 is the initial major semi-axis equal to
.1147020
kmhhRa AP =++=
Here R is the radius of the Earth, hP and hA are the altitude
values in the perigee and apogee. During the orbit evolution the
value of hP is changing a little. The radius of circular orbit will
be equal to
.6570 kmhRa PC = =+
With account of D=R, the time of orbit evolution is equal to
secaaFD
GMmT CC8
0 106.1)(2 ==
or about 5 years. 3. Problem. The magnitude of total umbral
lunar eclipse is equal to 1.865. Please find the duration of
totality. The expansion of the umbra caused by atmosphere can be
disregarded. (O.S. Ugolnikov) 3. . The value of eclipse magnitude
is too high. We have to define the distances between the Sun and
the Earth (L) and between the Earth and the Moon (l) for which that
value is possible. Lets assume that the eclipse is central and all
three bodies are situated along one line. Their position is shown
in the figure.
S
L
R
rl
Sun MoonEarth
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Russian Open School Astronomical Olympiad by Correspondence
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Lets define the umbra radius depending on L and r. The edge of
umbra cone tangent to Sun and Earth is inclined to the line
Sun-Earth by the angle
.L
RS=
This angle is close to the angular radius of the Sun, being
quite small (it is equal to 0.26) and we can use the trigonometric
properties of small angle. In particular, we can state that its
cosine is equal to unity. The umbra radius is equal to
.)( lSlLL
RlR = =
r
d
Moon
Umbra
As we can see in the figure, the central eclipse magnitude is
equal to
.22
12
1r
rr
rr
dF =+ +=+=
Here r is the radius of the Moon. It is clear that the more the
distance between the Sun and the Earth L, and the less the distance
between the Sun and the Moon, l, the more the eclipse magnitude. We
can see it using two initial formulae of solution. Lets assume that
the value of L is maximal (1.017 a.u.), so the Earth reaches the
aphelion point of the orbit. If we also assume that the value of l
is average (384,400 km) than the eclipse magnitude will not exceed
1.832, that will not met the condition of the problem.
Substituting the minimal value of l (356,400 km) we obtain the
maximal magnitude of the lunar eclipse (without account of
atmospheric expansion) equal to 1.868. It nearly coincides with the
one given in this problem. So, during the eclipse considered here
the Moon was near the perigee point of the orbit and crossed the
umbra along its diameter. The radius of the umbra is turned out to
be equal to 4757 km.
The lunar spatial velocity at the distance l is equal to
,12
=al
GMv
where a is the major semi-axis of the lunar orbit. Substituting
the numbers, we obtain 1.095 km/sec. The umbra also moves
relatively the Earth, its velocity is equal to
,0 Llvu =
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Russian Open School Astronomical Olympiad by Correspondence
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where v0 is the aphelion orbital velocity of the Earth (equal to
29.3 km/s). The value of u is equal to 0.069 km/s, the direction is
the same with the velocity of the Moon v. During the total eclipse
the Moon covers the distance
),(2 rD =
and the total eclipse duration is equal to
,)(2 rv vu u DT ==
or 5885 seconds or 1 hour and 38.1 minutes. Here we should
notice that it is not the maximal duration of the total lunar
eclipse that is reached during the apogee eclipses with smaller
magnitude.
Note. In the ephemeredes of the lunar eclipses published in
astronomical calendars and handbooks you can find the eclipses with
magnitudes higher than 1.868. This fact is due to the account of
atmospheric expansion of the umbra, which size is considered to be
more than the geometrical value. 4. Problem. The grazing
occultation of the star by the Moon is observed in the zenith at
the Earths equator. The Moon is exactly in the first quarter.
Please find the maximum possible angular distance between the star
being occulted and the closest horn of the Moon (the crossing point
of limb and terminator) in the grazing moment. The orbit of the
Moon can be considered to be circular. (O.S. Ugolnikov) 4.
Solution. Lets draw the positions of Sun, Moon, star and ecliptic
line during the occultation event. The cusps of the Moon are
directed along the major circle of the celestial sphere connecting
the Sun and the Moon. This circle is shown as the arc in the
figure. We have to note that due to the inclination of the lunar
orbit this circle does not coincide (in general) with ecliptic and
visual path of the Moon on the celestial sphere.
EclipticSun
Star
Moonr
Visual path of the Moon
90
Since the Sun is always situated on the ecliptic and the angular
distance between the Sun and the Moon in the first quarter is equal
to 90, cusps of the Moon are directed parallel to the ecliptic, not
depending on the position of the Moon relatively the ecliptic. If
the lunar visual motion relatively the star is directed parallel to
the ecliptic, than the grazing occultation will be observed exactly
at the horn point, the crossing of limb and terminator.
The angular distance between the star and the horn depends on
the inclination of the lunar path to the ecliptic . So we have to
define the maximal value of this angle. If we had observed from the
center of the Earth, the problem would be much simpler. The value
would be equal to the inclination of the lunar orbit, i, equal to
5.15. It would be reached, when the Moon had crossed the node of
its orbit. But really we observe from the surface of rotating Earth
and the value of can exceed the value of i.
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Russian Open School Astronomical Olympiad by Correspondence
2008
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According to the problem, the occultation is observed in the
zenith at the equator of the Earth. Since we have to find the
maximal value of angle , we assume that the Moon is situated on the
ecliptic crossing the orbit node. Let it be the ascending node.
Eclipticplane
Earth
Moon
Equator
i
L
R
vu
v
x
y
0
Figure shows the configuration of the Earth and the Moon on the
ecliptic plane. The Moon crosses this plane by the angle i and
moves northwards. We denote its velocity as v. Considering the
orbit as circular, we find this velocity:
.L
GMv =
Here M is the mass of the Earth, L the distance between the
Earth and the Moon. The velocity is equal to 1.02 km/sec. The
observer is situated on the equator and moves with it by the angle
(23.4) to the ecliptic. The velocity is equal to
.20R
Tv =
Here R is the radius of the Earth, T is the duration of sidereal
day. The value is equal to 0.465 km/s. The velocity of the Moon
relatively the observer is the vector difference of two velocities
above:
u = v v0.
The angle to find is the angle between vector u and the ecliptic
plane. This angle is maximal if the vertical components of vectors
v and v0 are opposite and the observer moves southwards. It is if
the Moon is seen in the vernal equinox point. The situation is the
same if the Moon is in descending angle and is seen in the autumn
equinox point.
We define the coordinate system (x, y) as shown in the figure
and make a projection of vector difference above on the axes of the
system:
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Russian Open School Astronomical Olympiad by Correspondence
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,coscos 0vivux = = + .sinsin 0vivu y The angle is equal to
.1.25coscossinsin
arctanarctan0
0 o
x
y
vivviv
uu =
+==
We have to note that this angle is almost 5 times larger than
for the case of geocentric observations. The angular distance
between the star and horn is equal to
.2
sin22
sin2 Lr==
Here is the angular radius of the Moon. Transforming this angle
to the degree scale, we obtain 0.113 or 6.8. If the grazing
occultation occurs at the dark edge of the Moon (as shown in the
first figure of solution), that angular distance will be enough to
observe the event using the binocular or telescope even in the case
of faint star. 5. Problem. The minor planet moves around the Sun in
the ecliptic plane, never coming inside the orbit of the Earth. The
conditions of its observations exactly repeat in 2 years, and its
visible magnitude changes on 8m with the same period. Please find
the minimum possible value of the eccentricity of the asteroids
orbit. The asteroid is the smooth spherical uniform ball with
constant surface albedo. Orbit of the Earth can be considered to be
circular. (O.S. Ugolnikov) 5. Solution. In two years the Earth
completes two revolutions around the Sun returning to the same
point of the orbit. Since the conditions of asteroid observations
are the same, it also returns to the same orbit point in two years.
Since it does not come inside the orbit of the Earth, its orbit
major semi-axis is not less than 1 a.u. and orbital period is not
less than 1 year. The number of completed revolutions in 2 years is
one or two.
But if this number was two and orbital period was equal to 1
year, asteroid would come inside the orbit of the Earth (in the
case of elliptical orbit) or would be in the fixed position
relatively the line Sun-Earth not changing the magnitude (in the
case of circular orbit). Possible axial rotation of the asteroid
does not change the picture, since the asteroid has the uniform
surface. Finally, the orbital period of asteroid is equal to 2
years. According to Third Kepler law, the major semi-axis of the
orbit is equal to 22/3 or 1.587 a.u.
Asteroid is the uniform ball with smooth surface. This case its
magnitude does not depend on the phase angle (the angle between the
directions from the asteroid to the Sun and the Earth). The
magnitude changes are related with the changes of distances between
the Sun and asteroid and between the asteroid and the Earth. The
asteroid brightness is reverse-proportional to the squares of both
distances and the magnitude can be expressed as follows:
m = m0 + 5 lg d + 5 lg r.
Here d and r are the distances from the asteroid to the Earth
and to the Sun expressed in astronomical units, m0 is the asteroid
absolute magnitude (the magnitude for the case d = r = 1 a.u.).
Let e be the eccentricity of asteroid orbit. In the case of
definite values of e and orbital period (two years) maximal
amplitude of brightness changes will be reached in the case shown
in the figure. During the opposition asteroid is in the orbit
perihelion, position 1 in the figure. The values of d and r reach
the minimum simultaneously:
d1 = a (1 e) a0, r1 = a (1 e).
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Russian Open School Astronomical Olympiad by Correspondence
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12Earth
Sun
Here a0 is the radius of Earths orbit. The brightness of the
asteroid reaches the maximal possible value. In one year the Earth
returns to the same point, but asteroid is in aphelion and in the
conjunction with the Sun. The values of d and r reach the maximum
simultaneously:
d2 = a (1 + e) + a0, r2 = a (1 + e).
The brightness of asteroid reaches the absolute minimal value.
Since the amplitude increases with the eccentricity, to find the
minimal value of e, we have to find for the configuration described
above for given amplitude 8m:
5 lg d2 + 5 lg r2 5 lg d1 5 lg r1 = 8,
.8.3910)1())1(()1())1(( 6.1
0
0
11
22 ==== Keaaeaeaaea
rdrd
The solution of square equation gives the answer:
.)1(2
16)1()1()1(2 2202
0
aKKaaKaKaK
e ++=
Just one of two solutions (with the sign ) has the physical
sense (0
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Russian Open School Astronomical Olympiad by Correspondence
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6. Solution. The coordinates of the stars change by two reasons:
light aberration and parallax of the stars. We will analyze these
effects separately.
Light aberration is caused by orbital motion of the Earth and
the finite value of the speed of light. It moves the star towards
the apex of the Earths motion. The ecliptic longitude of the apex
is 90 less than the one of the Sun. For the stars on the ecliptic
the change of ecliptic longitude is equal to:
).cos()sin( 0 llklavclA ==
Here v is the velocity of the Earth, c is the light velocity, k
is the aberration constant, equal to 20.5 in degree scale, a is the
apex ecliptic latitude, l0 is the ecliptic latitude of the Sun, l
is the ecliptic latitude of the star.
Parallax shift of the star is always directed towards the Sun.
The change of ecliptic longitude is equal to
).. sin( 0 lllP =
Here is the parallax of the star. Since the both shift values
are too small, the common shift is the sum of these values, and we
can us the observed values of star ecliptic longitude in these
formulae.
SunEarth 1
Earth 2 Apex
Apex
Y X
Y X
l
l ll l
l l l
AA
P P
PA AP
According to the problem formulation, in the moment 1:
l01 lX = 100, l01 lY = 70.
The values of ecliptic longitude shifts owing to aberration and
parallax are equal to:
'.'1.. . 4''5.0''6.3)sin ()cos( 01011 XXXX llllkl = = = ++++ .
.= = = ++ '.'8.6''2.0''0.7)sin()cos( 01011 YYYY llllkl
The difference of ecliptic longitudes (lY lX)1 in the moment 1
is equal to 30 and 10.9 less than heliocentric difference (lY lX).
Thus, the heliocentric difference of longitudes is equal to
300010.9. This is the answer for the second question of the
problem.
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Russian Open School Astronomical Olympiad by Correspondence
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To answer on the first question, we see that in three months
l02 lX = 190, l02 lY = 160.
According to this,
. . )sin ()cos( 02022 XXXX llllkl = = = +++ '.'1.20''1.0''2.20 .
.= = =++ )sin()cos( 02022 YYYY llllkl .'4.19''1.0''3.19 + +
The ecliptic longitude difference (lY lX)2 is 0.7 less than the
heliocentric value (lY lX) and equal to 300010.2. 7. Problem. In
March 1997 we saw the bright comet Hale-Bopp with magnitude 1.5m.
Being observed from Earth, the brightest inner part of the comets
tail had the length about 10 and width about 1. Imagine that the
same time the spaceship with astronauts arrived to the comet and
landed on its core at the side opposite relatively the Sun. Will
the astronauts see the stars in the sky when they come to the
surface of the core? (O.S. Ugolnikov) 7. Solution. To answer on the
question, we have to look how do the brightness characteristics of
expanded objects change when we fly to them or from them. If we
approach to comet Hale-Bopp in 2 times for example, it will be
brighter in 4 times and will have the angular square 4 times more
than before. The surface brightness (or the magnitude of the
angular square unit) will not change. If we come inside the
expanded object, its emission will be seen from the major part of
the sky, but the surface brightness will not be higher (actually it
will be lower) than the one observed from large distance.
The surface brightness of the tail of comet Hale-Bopp is equal
to 1.5m per 10 square degrees. Each square degree contains 36002
square angular seconds. The magnitude of one square second is equal
to
m = 1.5 + 2.5 lg (1036002) = 18.8.
When we land on the comet surface at the side where the tail is
visible, its surface brightness will be the same. But it is just
4-5m brighter than the moonless night sky on the Earth. The comet
night sky will be like evening sky during the nautical twilight.
The human eye will see the stars up to 4m in these conditions.
Finally, the astronauts will see the stars from the surface of the
core of comet Hale-Bopp. 8. Problem. The star has the surface
temperature 15000 K and the radius equal to 10 radii of the Sun.
During the last 100 years this star produces the uniform stellar
wind blowing with the velocity 20 km/s. This substance created the
shell of gas and dust around the star with optical depth equal to
0.2. Please calculate the radii of inner and outer visible edges of
the shell, find the dependence of the dust density in the shell on
the distance from the star. Please find the temperature at the
outer edge of the shell, mass of the shell and the mass loss rate
of the star. The dust particles have the radius equal to 1 mkm,
density equal to 3 g/cm3 and fusion temperature equal to 1500 K.
Consider that the mass of the gas is 200 times larger than the mass
of the dust, but light absorption is caused only by dust. (A.M.
Tatarnikov) 8. Solution. We know that the mass loss rate M is
constant. Lets divide the shell into the thin layers with the
thickness r. Masses of each layer are the same, we denote is as M.
The number of particles is also the same for all layers. If the
layer radius is R, than the density of the layer will be equal
to
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Russian Open School Astronomical Olympiad by Correspondence
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.4
)(2
2
==
RR
rRMR inin
Here Rin is the radius of the inner border of the shell and in
is the density there. Last formula is the dependency of the density
on the distance from the star. To find the temperature dependency,
we denote dust particle radius as a. The energy falling to the
particle is equal to the energy emitting by the particle to the
surrounding space. If we denote the radius and temperature of the
star as R* T*, than we will have
.44
4 4222
2*
4* Taa
RRT
=
The temperature of particle is equal to
.2
** R
RTT =
At the inner border of the shell the temperature must be equal
to the dust fusion temperature. This case the dust will not reflect
the emission closer to the star. Denoting the fusion temperature as
T0, we obtain
.2
2**
=
TTRRin
It is equal to 500 solar radii or 50 radii of the star. The
outer border radius is more simple to calculate:
,tv .Rout
Here v is the stellar wind velocity (20 km/sec), t is the time
of wind outflow (100 years). The radius is equal to 6.31010 km or
90000 solar radii or 9000 radii of the star. We see that outer
border radius is many times more than the inner border radius.
Physically it shows that the shell does create at the time t. The
temperature of the outer border is equal to
.1102
** KR
RTTout
out ==
To find the mass of the shell, we have to express the optical
depth of the shell. We assume that the particles absorb the
emission as the black balls with radius a. The probability of the
photon to be absorbed in the layer with radius R and thickness R,
or the optical depth of the layer, is equal to
.)()( 22
2 RaR
RnaRRnR inin
==
Here n(R) is the particle concentration at the distance R, nin
is their concentration at the inner border of the shell. Total
optical depth is the sum of the optical depths of all layers. It is
expressed as the integral
.11)( 22222
ininoutin
inin
R
R
inin
R
RRna
RRRnadRa
RRnRd
out
in
out
in
=
==
Here we take into account that the outer radius of the shell is
sufficiently larger than the inner radius. The value of optical
depth is known, and we can use this formula to calculate the
concentration nin. The number of particles in thin layer with
radius R and thickness R is equal to
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Russian Open School Astronomical Olympiad by Correspondence
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.4
44)()( 222 R
aRRRnRRRnRN ininin ===
This value does not depend on radius, since the mass outflow is
constant. Taking it into account and knowing that outer radius is
many times less than the inner one, we obtain
.4
2aRR
N outin=
To find the total mass, we remember that the mass of gas is K
times more than the mass of dust. If 0 is the dust density, than
the shell mass can be expressed as follows:
.3
1634 0
03 KaRRNKaM outin ==
It is equal to 41025 kg or 2105 of the solar mass. This mass was
released in 100 years, so the mass loss rate is equal to 41023 kg
or 2107 solar masses per year.
9. Problem. The gamma-ray bursts sometimes happen in the distant
galaxies. These are the short (about several seconds) bursts of
gamma-ray emission with average energy of the photon equal to 1
MeV. To be registered on the Earth, the flux of such photons must
be not less than 50 phot/(cm2s). The luminosity of the burst is
equal to 1049 ergs per second, this energy is released inside two
opposite cones with angle at the top equal to 10. The gamma-ray
bursts are registered on the Earth once a week. What is the
frequency of gamma-ray bursts in one definite galaxy? How much
times more or less bursts we would see, if the cones of their
emission were two times narrower? (M.E. Prokhorov) 9. Solution.
Lets find which part of the sphere is covered by two cones of
gamma-ray emission. These cones draw two circles with radius 5 or
0.087 radians. We denote this angle as . This angle is quite small
and we consider the circles as plane figures. The part of the
sphere covered by cones is equal to the ratio of circles and sphere
squares:
.004.024
2 22 === b
The energy of gamma-ray photon is equal to 1 MeV or 1.6106 ergs.
Thus, the gamma-ray source emits 61054 photons. We denote this
value as J0. We find the maximal distance to the source to be
observed by the device with sensitivity E:
.2
121;
400
20
EJ
bEJR
bRJE ===
Here we assume that the Earth is situated inside one of two
emission cones. Thus, we can detect the gamma-ray burst at the
distance up to 500 Mpc. The change of flux due to the Universe
expansion is not sufficient at this distance.
The average concentration of galaxies in the Universe is equal
to 0.01 pc3. We can obtain this value from the total number of
galaxies in the Universe (1010) and the size of the Universe (10
Gpc). The number of galaxies inside the sphere with radius R is
equal to
.105~63
4 623
03
==bEJnnRN
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Russian Open School Astronomical Olympiad by Correspondence
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14
We see the bursts with the frequency F equal to 1 burst per week
or 50 bursts per year in this number of galaxies. But we cannot see
all the bursts in these galaxies as we cannot always be inside the
emission cones. Total frequency of burst in these galaxies is equal
to
Fb
F =0
or 12500 bursts per year. The bursts frequency in one galaxy is
equal to
.23623
0
23
0
01
=
==
JE
nF
JE
nbF
NFF
or once in 400 years. We will see 1/250 of these bursts, so the
frequency of observable bursts in one galaxy is once in 100,000
years.
We see that this value in the formula is proportional to the
square root of b or proportional to the cone angular radius . If
the radius was two times less, the total frequency of the bursts in
one galaxy would be two times less. But the value of F0 is reverse
proportional to the square root of b, so the total number of
observable bursts would increase in two times!
It is easy to explain. If the cone radius was two times less, we
could see two times more distant gamma-ray bursts. They would
fulfill 8 times larger volume in the Universe. The probability to
be inside the cones would be 4 times less. So, the frequency of
gamma-ray burst observations would increase in two times.
10. Problem. We know that the temperature of Cosmic Microwave
Background in the direction with Galactic coordinates l = 264 and b
= 48 is maximal, being by T = 3.35 mK more than average value.
Please find the velocity of our Galaxy as a whole relatively the
Cosmic Microwave Background. (E.N. Fadeev) 10. Solution. Firstly we
have to find the velocity of Sun relatively the Cosmic Microwave
Background (CMB). The temperature change is related with the
Doppler effect:
.0cv=
Here 0 is the average (through the sky) wavelength of the CMB
maximum, is the one for the direction described above, v is the
velocity of the Sun relatively CMB. Since the CMB emission is
thermal, the wavelength of maximal emission is reverse proportional
to the temperature:
29.0)(T
cm =
The velocity is equal to
.0TTc
TTT
cv ==
Thus, the Sun moves relatively CMB with the velocity 368 km/sec
in the direction with galactic coordinates l = 264 and b = 48. But
the Sun moves relatively the center of the Galaxy with the velocity
v0 equal to 220 km/sec and directed to the point with galactic
coordinates l0 = 90 and b0 = 0. The vector of common Galaxy
velocity relatively CMB is equal to
u = v v0.
-
Russian Open School Astronomical Olympiad by Correspondence
2008
15
SunGalactic
plane l = 90l = 270
b
v
vu
0 We see that the galactic longitude l corresponding the vector
v is close to 270, and we can assume that all three vectors are in
the figure plane perpendicular to the galactic plane. This case the
value of the velocity u can be calculated as follows:
.)180cos(2 020
2 bvvvvu o += It is equal to 540 km/sec. If we take into account
the difference of l and 270, the answer will be the same with the
exactness of 1 km/sec.